The coordinates of the point $P$ on the line $2x+3y+1=0$, such that $|PA-PB|$ is maximum, where $A$ is $(2,0)$ and $B$ is $(0,2)$ is

 

Given equation of line is $2x+3y+1=0.$

Given points are $A\left(2,0\right), B\left(0,2\right).$

Let the coordinates of point $P$ is $\left(h,k\right)$.

Let $PAB$ forms a triangle as with any three points we can form a triangle.

In a triangle the difference between two sides should be less then the third side.

$\Rightarrow |PA-PB|<\left|AB\right|$

Let the three points $PAB$ are collinear.

Then the difference of $PA$ and $PB$ will be $AB.$

$\Rightarrow |PA-PB|=\left|AB\right|$

By combining above two statements we can say that $|PA-PB|\le \left|AB\right|$.

Thus, maximum value of $|PA-PB|$ is $\left|AB\right|$, which is possible only when $P,A,B$ are collinear.

The value of $AB=\sqrt{{\left(2-0\right)}^2+{\left(0-2\right)}^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2.}$

As maximum of $|PA-PB|$ is $\left|AB\right|=2\sqrt{2}$,

$\Rightarrow |PA-PB|=2\sqrt{2}$

The equation of line $AB$ will be

$$\begin{gathered} y-0=\frac{2-0}{0-2}\left(x-2\right) \\ \Rightarrow x+y=2 \end{gathered}$$

As $P,A,B$ are collinear, the point $P$ lies on $AB,$and also $P$ lies on $2x+3y+1=0.$

$\Rightarrow h+k=2, 2h+3k+1=0$

By solving above equations we get values of $h,k.$

As $h=2-k$ from first equation, by substituting it in second one we get,

$$\begin{gathered} \Rightarrow 2\left(2-k\right)+3k+1=4-2k+3k+1=5+k=0 \\ \Rightarrow k=-5. \end{gathered}$$

As $k=-5\Rightarrow h=2-\left(-5\right)=2+5=7$.

Then, the coordinates of point $P\left(7,-5\right)$.

Thus, the coordinates of point $P$ is $\left(7,-5\right)$ and maximum value of $|PA-PB|$ is $2\sqrt{2}.$

Therefore, the correct answer is option $2.$